
Two dice are thrown at the same time. Find the probability of getting
(i). Same number on both dice
(ii). Different number on both dice
Answer
434.7k+ views
Hint: In this question, we will proceed by writing all the possible cases occurring when two dice are thrown at the same time. Then we need to find out the cases that will give the same number on both dice and also for a different number on both dice. Then we need to use the probability formula to find the probability.
Formula used :
\[Probability = \dfrac{\text{Number of favourable outcomes}}{\text{Total number of possible outcomes}}\]
Complete step-by-step solution:
Given, two dice are thrown at the same time.
\[S\ = \{(1,1),(1,2)\ ,(1,3)\ ,(1,4)\ ,(1,5)\ ,(1,6),\ \]
\[\ (2,1)\ ,(2,2)\ ,(2,3)\ ,(2,4)\ ,(2,5)\ ,(2,6),\ \]
\[\ (3,1)\ ,(3,2)\ ,(3,3)\ ,(3,4)\ ,(3,5)\ ,(3,6),\ \]
\[\ (4,1)\ ,(4,2)\ ,(4,3)\ ,(4,4)\ ,(4,5)\ ,(4,6),\ \]
\[\ (5,1)\ ,(5,2)\ ,(5,3)\ ,(5,4)\ ,(5,5)\ ,(5,6),\ \]
\[\ (6,1)\ ,(6,2)\ ,(6,3)\ ,(6,4)\ ,(6,5)\ ,(6,6)\}\]
The probability of total outcomes is \[6^{2}\]
\[n\left( S \right) = \ 36\]
Let \[A\] be the event of getting same number on both sides
\[A\ = \ \left\{ \left( 1,1 \right),\ \left( 2,2 \right),\ \left( 3,3 \right),\ \left( 4,4 \right),\ \left( 5,5 \right),\ \left( 6,6 \right) \right\}\]
Now,
\[n\left( A \right) = \ 6\]
The probability,
\[P\left( A \right) = \dfrac{n\left( A \right)}{n\left( S \right)}\]
\[P\left( A \right) = \dfrac{6}{36}\]
By simplifying,
We get,
\[P\left( A \right) = \dfrac{1}{6}\]
Thus the probability of getting same number on both dices is \[\dfrac{1}{6}\]
Let \[B\] be the probability of getting a different number on both dice.
We have already found the probability of getting the same number on both dice. Now in order to find the probability of different number on both sides, we need subtract the probability of getting same number on both dice and \[1\]
\[P\left( B \right) = 1 – P\left( A \right)\]
\[P\left( B \right) = 1 - \dfrac{1}{6}\]
By simplifying,
We get,
\[P\left( B \right) = \dfrac{5}{6}\]
Thus the probability of getting different number on both dice is \[\dfrac{5}{6}\]
I). The probability of getting same number on both dices is \[\dfrac{1}{6}\]
II). The probability of getting different number on both dice is \[\dfrac{5}{6}\]
Note: We need to calculate the number of favourable and possible outcomes in each case to calculate the probability of each of the given events. We should also be careful that we don’t count the same event repeatedly by mistake. As the number of dice increases the total number of outcomes increases in multiplication of 6 like for three dice we will have $6\times 6\times 6 $ outcomes.
Formula used :
\[Probability = \dfrac{\text{Number of favourable outcomes}}{\text{Total number of possible outcomes}}\]
Complete step-by-step solution:
Given, two dice are thrown at the same time.
\[S\ = \{(1,1),(1,2)\ ,(1,3)\ ,(1,4)\ ,(1,5)\ ,(1,6),\ \]
\[\ (2,1)\ ,(2,2)\ ,(2,3)\ ,(2,4)\ ,(2,5)\ ,(2,6),\ \]
\[\ (3,1)\ ,(3,2)\ ,(3,3)\ ,(3,4)\ ,(3,5)\ ,(3,6),\ \]
\[\ (4,1)\ ,(4,2)\ ,(4,3)\ ,(4,4)\ ,(4,5)\ ,(4,6),\ \]
\[\ (5,1)\ ,(5,2)\ ,(5,3)\ ,(5,4)\ ,(5,5)\ ,(5,6),\ \]
\[\ (6,1)\ ,(6,2)\ ,(6,3)\ ,(6,4)\ ,(6,5)\ ,(6,6)\}\]
The probability of total outcomes is \[6^{2}\]
\[n\left( S \right) = \ 36\]
Let \[A\] be the event of getting same number on both sides
\[A\ = \ \left\{ \left( 1,1 \right),\ \left( 2,2 \right),\ \left( 3,3 \right),\ \left( 4,4 \right),\ \left( 5,5 \right),\ \left( 6,6 \right) \right\}\]
Now,
\[n\left( A \right) = \ 6\]
The probability,
\[P\left( A \right) = \dfrac{n\left( A \right)}{n\left( S \right)}\]
\[P\left( A \right) = \dfrac{6}{36}\]
By simplifying,
We get,
\[P\left( A \right) = \dfrac{1}{6}\]
Thus the probability of getting same number on both dices is \[\dfrac{1}{6}\]
Let \[B\] be the probability of getting a different number on both dice.
We have already found the probability of getting the same number on both dice. Now in order to find the probability of different number on both sides, we need subtract the probability of getting same number on both dice and \[1\]
\[P\left( B \right) = 1 – P\left( A \right)\]
\[P\left( B \right) = 1 - \dfrac{1}{6}\]
By simplifying,
We get,
\[P\left( B \right) = \dfrac{5}{6}\]
Thus the probability of getting different number on both dice is \[\dfrac{5}{6}\]
I). The probability of getting same number on both dices is \[\dfrac{1}{6}\]
II). The probability of getting different number on both dice is \[\dfrac{5}{6}\]
Note: We need to calculate the number of favourable and possible outcomes in each case to calculate the probability of each of the given events. We should also be careful that we don’t count the same event repeatedly by mistake. As the number of dice increases the total number of outcomes increases in multiplication of 6 like for three dice we will have $6\times 6\times 6 $ outcomes.
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